AIPMT 1991 Chemistry Question Paper with Answer and Solution

(B) The existence of a positively charged nucleus was established by Ernest Rutherford through his famous $\alpha$-particle scattering experiment (also known as the Geiger-Marsden experiment). In this experiment,he bombarded a thin gold foil with $\alpha$-particles and observed that a small fraction of particles were deflected at large angles,leading to the conclusion that the positive charge and mass of the atom are concentrated in a very small central region called the nucleus.

(B) Hund's Rule of Maximum Multiplicity states that for a given electron configuration,the term with the greatest multiplicity has the lowest energy. In simpler terms,every orbital in a subshell is singly occupied with one $e^-$ before any one orbital is doubly occupied,and all $e^-$ in singly occupied orbitals have the same spin.

(D) The Pauli Exclusion Principle states that in an atom or molecule,no two electrons can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
This principle implies that an orbital can hold a maximum of two electrons,and these two electrons must have opposite spins.

(D) The maximum number of electrons in a subshell is given by the formula $2(2l + 1)$.
For the azimuthal quantum number $l = 3$,the subshell is the $f$-subshell.
Substituting $l = 3$ into the formula: $2(2 \times 3 + 1) = 2(6 + 1) = 2(7) = 14$.
Therefore,the maximum number of electrons is $14$.

(B) According to the Aufbau principle,electrons fill orbitals in the order of increasing energy,which is determined by the $(n + l)$ rule.
For the given orbitals:
$4s: n=4, l=0 \implies n+l = 4$
$3d: n=3, l=2 \implies n+l = 5$
$4p: n=4, l=1 \implies n+l = 5$
$5s: n=5, l=0 \implies n+l = 5$
$4d: n=4, l=2 \implies n+l = 6$
Comparing the $(n+l)$ values and the $n$ values for orbitals with the same $(n+l)$,the correct order is $4s < 3d < 4p < 5s < 4d$.

(A) The geometry associated with each hybridisation is as follows:
$A. sp$ hybridisation results in a linear geometry.
$B. sp^2$ hybridisation results in a trigonal planar geometry.
$C. sp^3$ hybridisation results in a tetrahedral geometry,which is non-planar.
$D. dsp^2$ hybridisation results in a square planar geometry.
Therefore,$sp^3$ hybridisation is the one that results in non-planar orbitals.

(B) The bond angle is maximum for $sp$ hybridisation.
In $sp$ hybridisation,the two hybrid orbitals are arranged linearly with a bond angle of $180^\circ$.
In $sp^2$ hybridisation,the bond angle is $120^\circ$.
In $sp^3$ hybridisation,the bond angle is approximately $109.5^\circ$.
In $dsp^2$ hybridisation,the bond angle is $90^\circ$.

(C) The compound $X$ is carbon tetrachloride $(CCl_4)$.
$CCl_4$ undergoes $sp^3$ hybridization and possesses a perfectly symmetrical tetrahedral geometry because all four substituents attached to the central carbon atom are identical ($Cl$ atoms).
In molecules like chloromethane $(CH_3Cl)$,chloroform $(CHCl_3)$,or iodoform $(CHI_3)$,the presence of different atoms attached to the central carbon leads to unequal bond pairs and lone pair-bond pair repulsions,causing the bond angles to deviate from the ideal tetrahedral angle of $109^o 28'$.

(D) According to Boyle's law,for a given mass of an ideal gas at constant temperature,$V \propto \frac{1}{P}$.
This implies $V = \frac{k}{P}$,where $k$ is a constant.
Therefore,$PV = k$,meaning the product of pressure and volume always remains constant.

(C) In a closed flask,the volume $(V)$ and the mass of the gas remain constant.
According to Gay-Lussac's Law,at constant volume,$P \propto T$.
Since the temperature increases from $300 \ K$ to $600 \ K$,the pressure $(P)$ must increase.
As temperature increases,the average kinetic energy of the gas molecules increases,which leads to an increase in the frequency and rate of collisions.
Since the flask is closed,no gas is added or removed,so the number of moles $(n)$ remains constant.
Therefore,the statement that the number of moles of gas increases is incorrect.

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given: $P_1 = 1 \, atm$,$V_1 = 12000 \, L$,$T_1 = 27 + 273 = 300 \, K$
$P_2 = 0.5 \, atm$,$T_2 = -23 + 273 = 250 \, K$
$V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1} = \frac{1 \times 12000 \times 250}{0.5 \times 300} \, L$
$V_2 = \frac{12000 \times 250}{150} = 20000 \, L$
Therefore,the correct option is $(B)$.

(C) Let the mass of each gas be $m \, g$.
Number of moles of lighter gas $(n_1)$ $= \frac{m}{4}$.
Number of moles of heavier gas $(n_2)$ $= \frac{m}{40}$.
Total number of moles $(n_{total})$ $= \frac{m}{4} + \frac{m}{40} = \frac{10m + m}{40} = \frac{11m}{40}$.
Mole fraction of lighter gas $(x_1)$ $= \frac{n_1}{n_{total}} = \frac{m/4}{11m/40} = \frac{10}{11}$.
Partial pressure of lighter gas $= x_1 \times P_{total} = \frac{10}{11} \times 1.1 \, atm = 1 \, atm$.

(B) The number of moles of $CO_2$ present in $200 \ mL$ solution is calculated as:
$n = \text{Molarity} \times \text{Volume (in L)} = 0.1 \times \frac{200}{1000} = 0.02 \ mol$.
Since $1 \ mol$ of an ideal gas occupies $22.4 \ L$ at $STP$,the volume of $0.02 \ mol$ of $CO_2$ is:
$V = 0.02 \ mol \times 22.4 \ L/mol = 0.448 \ L$.
Thus,the correct option is $(B)$.

(B) The molecular weight of the gas is calculated as: $M = 2 \times \text{Vapour Density} = 2 \times 11.2 = 22.4 \ g/mol$.
At $N.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Since the molar mass is $22.4 \ g/mol$,$22.4 \ g$ of the gas occupies $22.4 \ L$.
Therefore,the volume occupied by $11.2 \ g$ of the gas is: $\frac{22.4 \ L}{22.4 \ g} \times 11.2 \ g = 11.2 \ L$.

(D) The ideal gas equation is given by $PV = nRT$.
Since the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we can write $PV = \frac{m}{M}RT$.
Rearranging the terms to solve for density $(d = \frac{m}{V})$,we get $d = \frac{m}{V} = \frac{PM}{RT}$.
Thus,the correct option is $D$.

(C) According to the kinetic molecular theory of gases,molecules of an ideal gas are in constant random motion and possess a distribution of speeds (Maxwell-Boltzmann distribution).
Therefore,it is not true that all molecules move with the same speed.

(B) $HCl$ is a strong acid and dissociates completely in water as follows:
$HCl \rightarrow H^+ + Cl^-$
Since the concentration of $HCl$ is $0.001 \, M$ (or $10^{-3} \, M$),the concentration of $H^+$ ions is $[H^+] = 10^{-3} \, M$.
The $pH$ is calculated using the formula $pH = -\log[H^+]$.
$pH = -\log(10^{-3}) = 3$.

(B) Given,$pH = 4$.
We know that at $25^{\circ}C$,$pH + pOH = 14$.
Therefore,$pOH = 14 - pH = 14 - 4 = 10$.
The concentration of hydroxide ions is given by $[OH^-] = 10^{-pOH}$.
Thus,$[OH^-] = 10^{-10} \ M$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$,the change in the number of gaseous moles is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting this value into the equation,we get: $\Delta H = \Delta E + (-2)RT = \Delta E - 2RT$.

(C) For an ideal gas,the internal energy $(U)$ and enthalpy $(H)$ are functions of temperature only.
During an isothermal process,the temperature remains constant,so $\Delta T = 0$.
Since $\Delta H = nC_p\Delta T$,if $\Delta T = 0$,then $\Delta H = 0$.
Therefore,the enthalpy remains unaffected.

(D) The ionization potential is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
Ionization potential generally decreases down a group as the atomic size increases and the valence electron is further from the nucleus.
Among the given options,$1s^1$ is Hydrogen,$1s^2, 2s^2, 2p^6$ is Neon (a noble gas with very high ionization potential),$1s^2, 2s^2, 2p^2$ is Carbon,and $1s^2, 2s^2, 2p^6, 3s^1$ is Sodium.
Sodium $(Na)$ is an alkali metal in the third period,and its valence electron is in the $3s$ orbital,which is much further from the nucleus compared to the valence electrons of the other elements listed.
Therefore,the configuration $1s^2, 2s^2, 2p^6, 3s^1$ corresponds to the lowest ionization potential.

(D) The correct answer is $(D)$.
Electronegativity generally increases across a period from left to right and decreases down a group.
Among the halogens $(F, Cl, Br, I)$,fluorine $(F)$ is at the top of the group.
Due to its smallest atomic size and highest effective nuclear charge,$F$ has the highest electronegativity among all elements in the periodic table.

(C) The correct option is $C$.
Ionization potential generally decreases down a group and increases across a period.
Option $A$ $(1s^1)$ is Hydrogen.
Option $B$ $(1s^2, 2s^2, 2p^6)$ is Neon,a noble gas with a very high ionization potential.
Option $C$ $(1s^2, 2s^2, 2p^6, 3s^1)$ is Sodium,an alkali metal $(Group \ 1)$. Alkali metals have the lowest ionization potential in their respective periods.
Option $D$ $(1s^2, 2s^2, 2p^2)$ is Carbon.
Comparing these,Sodium $(3s^1)$ has the lowest ionization potential due to its larger atomic size and lower effective nuclear charge compared to the others.

(D) When ammonia $(NH_3)$ is dissolved in water,it undergoes a reversible reaction to form ammonium hydroxide $(NH_4OH)$,which further dissociates into ammonium ions $(NH_4^+)$ and hydroxide ions $(OH^-)$.
The equilibrium is represented as: $NH_3 + H_2O \rightleftharpoons NH_4OH \rightleftharpoons NH_4^+ + OH^-$.
Therefore,the solution contains $NH_4^+$,$OH^-$,and undissociated $NH_4OH$ molecules.

(C) The structure is $(CH_3)_3C-CH=CH_2$.
$1$. Identify the longest carbon chain containing the double bond: The longest chain has $4$ carbon atoms,so the parent alkane is butane,and with the double bond,it is but$-1-$ene.
$2$. Number the chain starting from the end closer to the double bond: $C_1=CH_2$,$C_2=CH$,$C_3=C(CH_3)_2$,$C_4=CH_3$.
$3$. Identify substituents: There are two methyl groups at the $C_3$ position.
$4$. Combine the parts: $3,3-$dimethylbut$-1-$ene.

(B) The bond angle of ${109^o}28'$ is characteristic of a perfect tetrahedral geometry.
In $CCl_4$ (Carbon tetrachloride),the central carbon atom is $sp^3$ hybridized and is bonded to four identical chlorine atoms.
Due to the symmetry of the molecule,all bond angles are exactly ${109^o}28'$.
In other options like $CHCl_3$ or $CH_3Cl$,the presence of different atoms ($H$ and Cl) causes distortion in the bond angles due to different electronegativities and bond pair-bond pair repulsions.

(A) Step $1$: Dehalogenation of $2,3-$dibromo$-3-$methylpentane with zinc dust $(Zn)$ leads to the formation of an alkene via elimination of $ZnBr_2$.
$CH_3-CH(Br)-C(Br)(CH_3)-CH_2-CH_3 + Zn \to CH_3-CH=C(CH_3)-CH_2-CH_3 + ZnBr_2$.
Step $2$: The resulting alkene ($3-$methylpent$-2-$ene) is heated with $HI$ in the presence of red phosphorus $(Red \ P)$. Red phosphorus and $HI$ act as a strong reducing agent,which reduces the alkene to the corresponding alkane.
$CH_3-CH=C(CH_3)-CH_2-CH_3 + 2HI \xrightarrow{Red \ P} CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ ($3-$methylpentane).
Thus,the final product is $3-$methylpentane.

(B) In ethylene $(CH_2=CH_2)$,each carbon atom is bonded to two hydrogen atoms and one other carbon atom via a double bond.
Each carbon atom is surrounded by three electron domains (two single bonds and one double bond),which corresponds to $sp^2$ hybridisation.
The geometry around each carbon atom is trigonal planar.

(A) Baeyer's reagent is an alkaline solution of cold $KMnO_4$.
It is primarily used for the detection of unsaturation (double or triple bonds) in a molecule,as it causes the purple color of $KMnO_4$ to disappear.

(C) The bond length between carbon atoms depends on the bond order. As the bond order increases, the bond length decreases.
The order of bond lengths is: $C-C > C=C > C \equiv C$.
$1.$ Ethane $(C_2H_6)$ has a $C-C$ single bond $(154 \ pm)$.
$2.$ Ethene $(C_2H_4)$ has a $C=C$ double bond $(134 \ pm)$.
$3.$ Ethyne $(C_2H_2)$ has a $C \equiv C$ triple bond $(120 \ pm)$.
Therefore, the carbon-carbon bond length is minimum in Ethyne.

(D) The reaction of calcium carbide $(CaC_2)$ with water $(H_2O)$ is a standard laboratory method for the preparation of acetylene ($C_2H_2$ or $CH \equiv CH$).
The balanced chemical equation is:
$CaC_2 + 2H_2O \to CH \equiv CH + Ca(OH)_2$
Therefore,the correct option is $(D)$.

(D) The reaction of benzene with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ is known as the Friedel-Crafts acylation reaction.
In this reaction,the acetyl group $(-COCH_3)$ replaces a hydrogen atom on the benzene ring to form acetophenone $(C_6H_5COCH_3)$.
The chemical equation is: $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$.
Therefore,the correct option is $(d)$.

(D) Anhydrous $AlCl_3$ acts as a Lewis acid because it is an electron-deficient compound.
In the Friedel-Crafts reaction,it reacts with the alkyl halide to generate a carbocation (electrophile).
The reaction is: $CH_3Cl + AlCl_3 \to CH_3^+ + AlCl_4^-$

(C) Calculate the mass of each substance:
$1$. $25 \ g$ of mercury = $25 \ g$.
$2$. $2 \ moles$ of water $(H_2O)$: Molar mass = $(2 \times 1) + 16 = 18 \ g/mol$. Mass = $2 \times 18 = 36 \ g$.
$3$. $2 \ moles$ of carbon dioxide $(CO_2)$: Molar mass = $12 + (2 \times 16) = 44 \ g/mol$. Mass = $2 \times 44 = 88 \ g$.
$4$. $4 \ g$ atoms of oxygen $(O)$: Mass = $4 \times 16 = 64 \ g$.
Comparing the masses: $88 \ g > 64 \ g > 36 \ g > 25 \ g$.
Therefore,$2 \ moles$ of carbon dioxide is the heaviest. The correct option is $C$.

(A) $BaSO_4$ has a very high lattice energy due to the strong electrostatic forces between the $Ba^{2+}$ and $SO_4^{2-}$ ions.
Since the hydration energy released upon dissolution is not sufficient to overcome this high lattice energy,$BaSO_4$ exhibits low solubility in water.

(B) Atoms of different elements having different atomic numbers but the same mass number are called isobars.

(B) The conversion of propene to $1-propanol$ is an anti-Markovnikov hydration reaction,which is achieved via hydroboration-oxidation.
Step $1$: Hydroboration: $3CH_3-CH=CH_2 + \frac{1}{2}B_2H_6 \rightarrow (CH_3-CH_2-CH_2)_3B$
Step $2$: Oxidation: $(CH_3-CH_2-CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3-CH_2-CH_2-OH + B(OH)_3$
Thus,the reagents $B_2H_6$ and alkaline $H_2O_2$ are ideal for this conversion.

(A) The reaction sequence is as follows:
$1$. $C_2H_2$ (acetylene) undergoes hydration in the presence of $HgSO_4$ and $H_2SO_4$ to form acetaldehyde $(CH_3CHO)$,which is $A$.
$2$. Acetaldehyde $(CH_3CHO)$ on oxidation $([O])$ yields acetic acid $(CH_3COOH)$,which is $B$.
$3$. Therefore,$B$ is an acid.

(C) The juice of the lemon contains approximately $5\%$ to $6\%$ citric acid.
This acid provides the characteristic sour taste and results in a $pH$ value of around $2.2$.

(A) Ferredoxin is an iron-sulfur protein that acts as an electron carrier in the photosynthetic electron transport chain.
It is specifically associated with $PS-I$ (Photosystem $I$).
During the light-dependent reactions,electrons are transferred from $PS-I$ to ferredoxin,which then reduces $NADP^+$ to $NADPH$ via the enzyme ferredoxin-$NADP^+$ reductase.

(A) Abscisic Acid $(ABA)$ is known as a stress hormone in plants.
It is synthesized during periods of water stress or drought.
$ABA$ triggers the closure of stomata to reduce the rate of transpiration,thereby helping the plant conserve water.

(C) dihybrid condition refers to an organism that is heterozygous for two different traits or genes.
In the given options,$Tt\, Rr$ represents an individual who is heterozygous for both the $T$ (tall/dwarf) gene and the $R$ (round/wrinkled) gene.
$tt\, rr$ is a homozygous recessive individual.
$Tt\, rr$ is heterozygous for one trait and homozygous for the other.
$TT\, Rr$ is homozygous for one trait and heterozygous for the other.
Therefore,$Tt\, Rr$ is the correct representation of a dihybrid.

(B) dihybrid condition refers to an organism that is heterozygous for two different traits or genes.
In the given options,the genotype $TtRr$ represents an individual that is heterozygous for both the height trait $(Tt)$ and the seed shape/color trait $(Rr)$.
Therefore,$TtRr$ is the correct representation of a dihybrid genotype.

(B) The energy of a photon is given by $E = hf = \frac{hc}{\lambda}$.
Given,$E_1 = 1 \ keV = 10^3 \ eV$ and $E_2 = 1 \ MeV = 10^6 \ eV$.
The frequency $f$ is directly proportional to the energy $E$ $(f = E/h)$.
Therefore,$\frac{f_2}{f_1} = \frac{E_2}{E_1} = \frac{10^6 \ eV}{10^3 \ eV} = 10^3$.
First,find the frequency $f_1$ of the $1 \ keV$ photon:
$f_1 = \frac{c}{\lambda_1} = \frac{3 \times 10^8 \ m/s}{1.24 \times 10^{-9} \ m} \approx 2.419 \times 10^{17} \ Hz$.
Now,calculate the frequency $f_2$ of the $1 \ MeV$ photon:
$f_2 = f_1 \times 10^3 = 2.419 \times 10^{17} \times 10^3 \approx 2.4 \times 10^{20} \ Hz$.

(A) According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $V_1 = 2 \ L$,$P_2 = 2P_1$,and $T_2 = 2T_1$.
Substituting the values into the equation:
$\frac{P_1 \times 2 \ L}{T_1} = \frac{(2P_1) \times V_2}{(2T_1)}$.
Solving for $V_2$:
$V_2 = \frac{P_1 \times 2 \ L \times 2T_1}{T_1 \times 2P_1} = 2 \ L$.
Thus,the volume remains $2 \ L$.

(B) The root mean square speed $(V_{rms})$ is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass of the gas.
Since $V_{rms} \propto \frac{1}{\sqrt{M}}$,the gas with the lowest molar mass will have the highest $V_{rms}$.
The molar masses are: $M(H_2) = 2 \ g/mol$,$M(N_2) = 28 \ g/mol$,$M(O_2) = 32 \ g/mol$,and $M(HBr) = 81 \ g/mol$.
Comparing the values: $\frac{1}{\sqrt{2}} > \frac{1}{\sqrt{28}} > \frac{1}{\sqrt{32}} > \frac{1}{\sqrt{81}}$.
Therefore,the order of $V_{rms}$ is $HBr < O_2 < N_2 < H_2$.

(C) The electronic configurations of the elements are: $C (2s^2 2p^2)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$,$F (2s^2 2p^5)$.
After the removal of the first electron,the configurations become: $C^+ (2s^2 2p^1)$,$N^+ (2s^2 2p^2)$,$O^+ (2s^2 2p^3)$,$F^+ (2s^2 2p^4)$.
The second ionization potential involves removing an electron from these cations.
$O^+$ has a stable half-filled $2p^3$ configuration,making it the most difficult to remove an electron from.
Following the periodic trend and stability,the order of second ionization potential is $O > F > N > C$.

(C) The correct answer is $C$ ($sp$ hybridization).

Hybridization Type	Bond Angle
$sp^3$	$109.5^\circ$
$sp^2$	$120^\circ$
$sp^3d$	$90^\circ$ and $120^\circ$
$sp$	$180^\circ$

As shown in the table,$sp$ hybridization corresponds to a linear geometry with a bond angle of $180^\circ$,which is the highest among the given options.

(B) When a $p-n$ junction is formed,the concentration gradient of charge carriers (holes in the $p$-region and electrons in the $n$-region) causes them to diffuse across the junction.
As electrons diffuse from the $n$-side to the $p$-side and holes diffuse from the $p$-side to the $n$-side,they recombine near the junction.
This recombination leaves behind immobile ionized impurity atoms (negative ions on the $p$-side and positive ions on the $n$-side).
These immobile ions create an electric field that opposes further diffusion,resulting in a region depleted of mobile charge carriers,known as the depletion layer.

(B) When a $p-n$ junction is formed,there is a concentration gradient of charge carriers across the junction. Holes diffuse from the $p$-side to the $n$-side,and electrons diffuse from the $n$-side to the $p$-side. As these charge carriers cross the junction,they recombine near the junction interface. This recombination leaves behind immobile ionized impurity atoms (negative ions on the $p$-side and positive ions on the $n$-side). This region,which is depleted of mobile charge carriers,is known as the depletion layer.

(B) The atomic number of copper is $29$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^9 4s^2$.
However,a fully-filled $3d$ subshell $(3d^{10})$ is more stable than a partially-filled one.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital to achieve the stable configuration $[Ar] 3d^{10} 4s^1$.

(D) The anhydride of an acid is formed by the removal of water molecules from the acid.
For nitric acid $(HNO_3)$,the reaction is:
$2HNO_3 \rightarrow N_2O_5 + H_2O$
Thus,$N_2O_5$ is the anhydride of $HNO_3$.

(B) The thermal decomposition of ammonium nitrite $(NH_4NO_2)$ yields pure $N_2$ gas.
$NH_4Cl_{(aq)} + NaNO_{2(aq)} \to NH_4NO_{2(aq)} + NaCl_{(aq)}$
$NH_4NO_{2(aq)} \xrightarrow{\Delta} N_{2(g)} + 2H_2O_{(l)}$
Note: While $(NH_4)_2Cr_2O_7$ also produces $N_2$,the reaction with $NH_4Cl$ and $NaNO_2$ is the standard laboratory method for preparing pure $N_2$.

(D) The correct option is $(d)$.
Concentrated nitric acid $(HNO_3)$ acts as a strong oxidizing agent.
It oxidizes cane sugar $(C_{12}H_{22}O_{11})$ to oxalic acid $((COOH)_2)$ and water.
The chemical equation for the reaction is:
$C_{12}H_{22}O_{11} + 18HNO_3 \rightarrow 6(COOH)_2 + 9NO_2 + 9H_2O$

(A) The reaction between phosphonium iodide $(PH_4I)$ and sodium hydroxide $(NaOH)$ is a base-acid reaction that produces phosphine gas $(PH_3)$,water $(H_2O)$,and sodium iodide $(NaI)$.
The balanced chemical equation is:
$PH_4I + NaOH \rightarrow PH_3 + H_2O + NaI$

(D) $P_2O_5$ reacts with water to form orthophosphoric acid.
The chemical equation is: $P_2O_5 + 3H_2O \to 2H_3PO_4$.
Therefore,the correct option is $D$.

(B) The hydrolysis of phosphorus trichloride $(PCl_3)$ with water proceeds as follows:
$PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
Thus,$PCl_3$ reacts with water to form phosphorous acid $(H_3PO_3)$ and hydrogen chloride $(HCl)$.

(C) $Oleum$ is also known as fuming $H_2SO_4$. It is prepared by dissolving $SO_3$ in $H_2SO_4$. Its chemical formula is $H_2S_2O_7$.

(B) Molarity is defined as the number of moles of solute dissolved per litre of solution.
Molarity = $\frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
Therefore,the unit of Molarity is $moles/litre$ (or $M$ or $mol \ L^{-1}$).
Molarity is temperature-dependent because volume changes with temperature,whereas Molality is temperature-independent.

(C) The molarity $(M)$ is calculated using the formula: $M = \frac{\text{mass of solute (g)} \times 1000}{\text{molar mass} \times \text{volume of solution (mL)}}$.
Given: mass of $Na_2SO_4 = 7.1 \ g$,molar mass $= 142 \ g/mol$,volume $= 100 \ mL$.
Substituting the values: $M = \frac{7.1 \times 1000}{142 \times 100} = \frac{7100}{14200} = 0.5 \ M$.

(B) According to Raoult's law for a dilute solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P^o - P}{P^o} = X_{solute} = \frac{n}{n + N}$.
Here,$n$ is the number of moles of solute and $N$ is the number of moles of solvent.
Therefore,the relative lowering of vapour pressure is equal to the ratio of the number of solute molecules to the total number of molecules in the solution $(n + N)$.

(A) An $Isotonic$ solution is one in which the osmotic pressure is equal to that of the blood plasma.
In such a solution,the water concentration outside the cell is the same as the water concentration within the cell.
Consequently,there is no net movement of water across the cell membrane,allowing the blood cells to retain their normal form.

(C) Osmotic pressure is a colligative property,which depends on the number of solute particles in the solution. The formula is $\pi = iCRT$,where $i$ is the van't Hoff factor.
For $M/10$ $HCl$,$i = 2$ $(H^+ + Cl^-)$.
For $M/10$ urea,$i = 1$ (non-electrolyte).
For $M/10$ $BaCl_2$,$i = 3$ $(Ba^{2+} + 2Cl^-)$.
For $M/10$ glucose,$i = 1$ (non-electrolyte).
Since $M/10$ $BaCl_2$ has the highest van't Hoff factor $(i = 3)$,it produces the maximum number of particles and thus has the highest osmotic pressure.

(D) $ (d) $ Crystals show good cleavage because their constituent particles are arranged in planes. Cleavage is the property of a crystal to break along specific planes of weakness.

(A) Cinnabar is the most common and is the chief ore of mercury.
Its chemical formula is $HgS$.

(C) Generally,$d$-block elements are called transition elements because they contain an inner partially filled $d$-subshell.
Thus,their general electronic configuration is represented as $(n - 1)d^{1 - 10}ns^{1 - 2}$.

(A) The atomic number of $Cr$ is $24$.
The expected electronic configuration is $[Ar] 3d^4 4s^2$.
However,the actual electronic configuration is $[Ar] 3d^5 4s^1$.
This configuration is more stable because of the symmetrical distribution of electrons in the $d$-orbitals and higher exchange energy associated with the half-filled $d$-subshell.

(B) When chloral $(CCl_3CHO)$ is boiled with an aqueous solution of sodium hydroxide $(NaOH)$,it undergoes a haloform reaction to produce chloroform $(CHCl_3)$ and sodium formate $(HCOONa)$.
The chemical equation is: $CCl_3CHO + NaOH \xrightarrow{\text{Boil}} CHCl_3 + HCOONa$.

(A) Grignard reagents $(RMgX)$ are strong nucleophiles and strong bases. They react readily with acidic protons,such as those found in the $ -COOH$ group,to form an alkane and a carboxylate salt. However,a simple $ -CH_3$ group (methyl group),a magnesium atom,or a halogen atom (when part of an alkyl halide) do not possess acidic protons or electrophilic centers that would undergo a characteristic reaction with a Grignard reagent in the same way. Among the given options,the methyl group $( -CH_3)$ is chemically inert towards Grignard reagents.

(A) In the given reaction,the hydroxide ion $(OH^-)$ acts as a nucleophile.
It attacks the electrophilic carbon atom bonded to the halogen $(X)$ and replaces the halide ion $(X^-)$.
Since a nucleophile replaces another nucleophile (the leaving group),this reaction is classified as a nucleophilic substitution reaction.

(B) Glycerol reacts with $P_4 + I_2$ (which generates $PI_3$ in situ) to initially form $1,2,3-triiodopropane$.
This intermediate is unstable due to the presence of iodine atoms on adjacent carbons and undergoes elimination of $I_2$ to form allyl iodide $(CH_2=CH-CH_2I)$.
$CH_2OH-CHOH-CH_2OH$ $\xrightarrow{P_4/I_2} [CH_2I-CHI-CH_2I]$ $\xrightarrow{-I_2} CH_2=CH-CH_2I$

(A) The Kolbe-Schmidt reaction is a base-promoted carboxylation of phenols.
In this reaction,sodium phenoxide is treated with carbon dioxide $(CO_2)$ under pressure,followed by acidification to yield salicylic acid as the final product.

(B) . The solubility of lower alcohols like methanol $(CH_3OH)$ and ethanol $(C_2H_5OH)$ in water is due to their ability to form intermolecular hydrogen bonds with water molecules.
The oxygen atom of the alcohol group carries a partial negative charge $(\delta-)$ and the hydrogen atom carries a partial positive charge $(\delta+)$,allowing them to interact with the water molecules through hydrogen bonding: $R-O^{\delta-}-H^{\delta+} \dots O^{\delta-}(H)-H^{\delta+}$.
Thus,option $B$ is correct.

(C) The reaction of an acid chloride with $H_2$ in the presence of $Pd-BaSO_4$ is known as the Rosenmund reduction.
This reaction specifically reduces the acid chloride group $(-COCl)$ to an aldehyde group $(-CHO)$.
Therefore,the product $P$ is $RCHO$.

(D) Acetaldehyde contains a carbonyl group $(C=O)$.
Due to the difference in electronegativity between carbon and oxygen,the carbonyl carbon is electrophilic,making it susceptible to attack by nucleophiles (nucleophilic addition reactions).
Additionally,the oxygen atom has lone pairs and can act as a weak nucleophile,and the alpha-hydrogens can be involved in reactions with electrophiles (e.g.,in aldol condensation or halogenation).
Therefore,acetaldehyde can react with both electrophiles and nucleophiles.

(A) The acidity of carboxylic acids is influenced by the inductive effect of substituents attached to the alpha-carbon.
Electron-withdrawing groups like $Cl$ increase the acidity by stabilizing the carboxylate anion through the $-I$ effect.
As the number of chlorine atoms increases,the $-I$ effect increases,thereby increasing the acidic strength.
The order of acidic strength is: $CH_3COOH < ClCH_2COOH < Cl_2CHCOOH < Cl_3CCOOH$.
Therefore,$CH_3COOH$ is the weakest acid.

(A) $CH_2=CH_2$ (ethylene) is a monomer.
Large number of ethylene monomeric units combine together through polymerization to form a polymer called polyethylene,as shown in the reaction: $n(CH_2=CH_2) \rightarrow -[CH_2-CH_2]_n-$.

(B) Natural rubber is a linear polymer of isoprene ($2$-methyl-$1,3$-butadiene).
Its monomer unit $X$ is isoprene,which has the chemical formula $CH_2=C(CH_3)-CH=CH_2$.

(A) Starch is a polysaccharide,which is a polymer of $D$-glucose units.
Upon complete hydrolysis,the glycosidic linkages in starch are broken down,resulting in the formation of $D$-glucose molecules only.

(B) Insulin is a protein consisting of $51$ amino acids arranged in two chains,$\alpha$ and $\beta$.
The $\alpha$-chain contains $21$ amino acids,and the $\beta$-chain contains $30$ amino acids.
It is secreted by the pancreas to regulate the blood sugar level.